The area (in sq. units) bounded by the parabola y2 = 8x and the line…

y² = 8x

so $y = 2\sqrt{2}\sqrt{x}$

x > 2       area in 1^st quadrant = area of 4^th quadrant

⇒  area in 1^st quadrant × 2

$=2 \times \int\limits_0^2 2 \sqrt{2} \sqrt{x} d x$

$=4 \sqrt{2}\left[\frac{2 x^{3 / 2}}{3}\right]_0^2=\frac{4 \sqrt{2} \times 2 \times 2 \sqrt{2}}{3}$

⇒  $\frac{32}{3}$ sq. units